Question

What are values of ΔH, ΔU, ΔS for 1 mole of H20?

Find ΔH, ΔU, ΔS for 1 mole of H20(l) on T=293K, p=101 325Pa and Cp(l)=75,36 J/molK and for 1 mole of H20(g) when T=520K, p=101 325 Pa, Cp(g)=36kJ/molK and
what is the change in enthalpy of vaporization of water on T=373K and p=101 325Pa and ΔH=40695,7 J/mol

NOTE: `C_P(g) = 36` `J"/"molcdotK`. Seems to be a typo.
- Truong-Son

Answer 1

The way the question is worded is quite confusing, but if that's the way it's worded, here's how I'm interpreting this.

Since the definitions of the three thermodynamic functions given are:

`DeltaH_(H_2O(l)) = int_(T_1)^(T_2) C_P(l)dT`
`DeltaS_(H_2O(l)) = int_(T_1)^(T_2) (C_P(l))/TdT`
`DeltaU_(H_2O(l)) = DeltaH_(H_2O(l)) - Delta(PV)`

we can work from there.

It seems reasonable (though tedious) that the question wants us to calculate for `"293 K" -> "373 K"`, through the vaporization process, and on to the heating process of `"373 K" -> "520 K"`, actually...

This would illustrate the definitions given above.

CHANGE IN ENTHALPY OF HEATING LIQUID WATER, VAPORIZING IT, THEN HEATING THE VAPOR

`DeltaH_(H_2O(l)) = int_("293 K")^("373 K") C_P(l)dT`

In this temperature range, we may assume that `C_P(l)` is approximately constant, so that:

`color(blue)(DeltaH_(H_2O(l))^(293 -> "373 K")) ~~ C_P(l)int_("293 K")^("373 K") dT`

`= ("75.36 J/mol"cdot"K")("373 - 293 K")`

`=` `color(blue)("6028.8 J/mol")`

Include the `DeltaH_"vap"` of `color(blue)("40695.7 J/mol")` for the vaporization process.

Next:

`color(blue)(DeltaH_(H_2O(g))^(373 -> "520 K")) ~~ C_P(g)int_("373 K")^("520 K") dT`

`= ("36 J/mol"cdot"K")("520 - 373 K")`
(not `36000`; the kilo was a typo.)

`=` `color(blue)("5292 J/mol")`

This gives a total `DeltaH` for the heating, vaporization, and further heating process as:

`bbcolor(blue)(nDeltaH) = n(DeltaH_(H_2O(l))^(293 -> "373 K") + DeltaH_"vap" + DeltaH_(H_2O(g))^(373 -> "520 K"))`

`= ("1 mol")(6028.8 + 40695.7 + 5292) = "52016.5 J"`

`=` `bbcolor(blue)("52.017 kJ")`

CHANGE IN ENTROPY OF HEATING LIQUID WATER, VAPORIZING IT, THEN HEATING THE VAPOR

Similarly, the change in entropy over a temperature range can be calculated by the same kind of approximation about `C_P(l)` and `C_P(g)`:

`color(blue)(DeltaS_(H_2O(l))^(293 -> "373 K")) ~~ C_P(l)int_("293 K")^("373 K") 1/TdT`

`= ("75.36 J/mol"cdot"K")ln(373/293)`

`=` `color(blue)("18.19 J/mol"cdot"K")`

At constant temperature and pressure, which is how it is at the phase equilibrium we establish for vaporization, `DeltaG = 0`, so that `DeltaH - TDeltaS = 0`. Therefore:

`color(blue)(DeltaS_"vap") = (DeltaH_"vap")/T_b = ("40695.7 J/mol")/("373 K")`

`=` `color(blue)("109.10 J/mol"cdot"K"`

Finally, a similar process for `DeltaS` going from `"373 K"` to `"520 K"` would be:

`color(blue)(DeltaS_(H_2O(g))^(373 -> "520 K")) ~~ C_P(g)int_("373 K")^("520 K") 1/TdT`

`= ("36 J/mol"cdot"K")ln(520/373)`

`=` `color(blue)("11.96 J/mol"cdot"K")`

So, the total `DeltaS` for this heating process on `"1 mol"` of water would be:

`bbcolor(blue)(nDeltaS) = n(DeltaS_(H_2O(l))^(293 -> "373 K") + DeltaS_"vap" + DeltaS_(H_2O(g))^(373 -> "520 K"))`

`= ("1 mol")(18.19 + 109.10 + 11.96)` `"J/mol"cdot"K"`

`=` `bbcolor(blue)("139.25 J/K")`

CHANGE IN INTERNAL ENERGY FOR HEATING LIQUID WATER, VAPORIZING IT, THEN HEATING THE VAPOR

Recall that

`DeltaH = DeltaU + Delta(PV)`

`= DeltaU + PDeltaV + VDeltaP + DeltaPDeltaV`.

Well, we've been at constant pressure, so we don't have to worry about the `VDeltaP + DeltaPDeltaV` term.

For the heating process, we'd need the densities of water at both temperatures to get accurate molar volumes (which are necessary because liquids are fairly incompressible). I think we can assume that `DeltaU ~~ DeltaH` for this first part because `PDeltaV` is going to be small.

But just to see...

`rho_("293 K") = "998.21 g/L"`
`rho_("373 K") = "958.37 g/L"`

`barV_("293 K") = M_m/rho = "g"/"mol" xx "L"/"g" = 18.015/998.21 = "0.018047 L/mol"`

`barV_("373 K") = M_m/rho = 18.015/958.37 = "0.018798 L/mol"`

So for `"1 mol"` of liquid water, `DeltaV = 0.018798 - 0.018047 = "0.000751 L"`, and so, from:

`DeltaH = DeltaU + Delta(PV) = DeltaU + PDeltaV`,
(at constant pressure)

`color(blue)(nDeltaU_(H_2O(l))^(293 -> "373 K")) = nDeltaH_(H_2O(l))^(293 -> "373 K") - PDeltaV_(H_2O(l))^(293 -> "373 K")`

`=` `("1 mol")("6028.8 J/mol") - ("1.01325 bar")("0.000751 L")xx("8.314472 J")/("0.0831345 L"cdot"bar")`

`=` `color(blue)("6028.7 J") ~~ DeltaH_(H_2O(l))^(293 -> "373 K")`

Indeed, the `PDeltaV` term hardly mattered for liquid water.

For the vaporization, a similar idea follows in that `DeltaV_((l)->(g)) ~~ V_((g))`. At high temperatures, we can treat water approximately as an ideal gas, and so:

`PDeltaV_((l)->(g)) ~~ PV_((g)) = nRT`

`= ("1 mol")("8.314472 J/mol"cdot"K")("373 K") = "3101.30 J"`

Therefore:

`color(blue)(nDeltaU_"vap") = nDeltaH_"vap" - PDeltaV_((l)->(g))`

`= ("1 mol")("40695.7 J/mol") - ("3101.30 J")`

`=` `color(blue)("37594.4 J")`

Finally, for the heating of the gas, again, we can assume ideality to get the change in volume of the water vapor. At constant `P` and `n`:

`PDeltaV = nRDeltaT`

`= ("1 mol")("8.314472 J/mol"cdot"K")("520 - 373 K")`

`=` `"1222.23 J"`

So:

`color(blue)(nDeltaU_(H_2O(g))^(373 -> "520 K")) = nDeltaH_(H_2O(g))^(373 -> "520 K") - PDeltaV`

`= ("1 mol")("5292 J/mol") - ("1222.23 J")`

`=` `color(blue)("4069.77 J")`

So, the overall `DeltaU` is:

`bbcolor(blue)(nDeltaU) = n(DeltaU_(H_2O(l))^(293 -> "373 K") + DeltaU_"vap" + DeltaU_(H_2O(g))^(373 -> "520 K"))`

`= ("1 mol")(6028.7 + 37594.4 + 4069.77)` `"J/mol"`

`=` `bbcolor(blue)("47.693 kJ")`